Green–Kubo formula for Boltzmann and Fermi–Dirac statistics

Abstract

Shear viscosity of nuclear matter is extracted via the Green–Kubo formula and the Gaussian thermostated SLLOD algorithm (the shear rate method) in a periodic box by using an improved quantum molecular dynamic (ImQMD) model without mean field, also it is calculated by a Boltzmann-type equation. Here a new form of the Green–Kubo formula is put forward in the present work. For classical limit at nuclear matter densities of $0.4\rho _{0}$ and $1.0\rho _{0}$, shear viscosity by the traditional and new form of the Green–Kubo formula as well as the SLLOD algorithm are coincident with each other. However, for non-classical limit, shear viscosity by the traditional form of the Green–Kubo formula is higher than those obtained by the new form of the Green–Kubo formula as well as the SLLOD algorithm especially in low temperature region. In addition, shear viscosity from the Boltzmann-type equation is found to be less than that by the Green–Kubo method or the SLLOD algorithm for both classical and non-classical limits.

Appendices

Appendix A: The Green–Kubo formula for shear viscosity

The particle density can be written as (for simplicity, here we use $\delta$-function to replace the Gaussian wave-packet). And the derivation of the Green–Kubo formula for shear viscosity is tedious. Here we give a simple introduction which is based on Ref. [64]. For more details one can find in Refs. [64, 68].

A.1 $\vec {\mathbf{r}}$ and $\vec {\mathbf{k}}$-space representations

The streaming velocity, mass density, momentum density and stress tensor at position $\vec {\mathbf{r}}$ and time t can be written:

$$\begin{aligned}&\mathbf{u} (\vec {\mathbf{r}},t)=\frac{\sum _{i}^{N} m_{i} {\dot{\vec {\mathbf{r}}}_{ {i}}} \delta (\vec {\mathbf{r}}-\vec {\mathbf{r}}_{i})}{\sum _{i}^{N} m_{i} \delta (\vec {\mathbf{r}}-\vec {\mathbf{r}}_{i})},\end{aligned}$$

(A1)

$$\begin{aligned}&\rho (\vec {\mathbf{r}},t)=\sum _{i}^{N} m_{i} \delta (\vec {\mathbf{r}}-\vec {\mathbf{r}}_{i}), \end{aligned}$$

(A2)

$$\begin{aligned}&J(\vec {\mathbf{r}},t)=\rho (\vec {\mathbf{r}},t)\mathbf{u} (\vec {\mathbf{r}},t)=\sum _{i}^{N} m_{i} {\dot{\vec {\mathbf{r}}}_{ {i}}} \delta (\vec {\mathbf{r}}-\vec {\mathbf{r}}_{i}), \\ P_{\alpha \beta }(\vec {\mathbf{r}},t)&=\sum _{i}^{N} \frac{p_{{i}\alpha }p_{{i}\beta }}{m_{i}} \delta (\vec {\mathbf{r}}-\vec {\mathbf{r}}_{i}) \nonumber \end{aligned}$$

(A3)

$$\begin{aligned}&+\frac{1}{2} \sum _{i}^{N} \sum _{i\ne j}^{N} F_{ij\alpha }R_{ij\beta } \delta (\vec {\mathbf{r}}-\vec {\mathbf{r}}_{j}) \\ \nonumber&+\ldots , \end{aligned}$$

(A4)

where ‘i’ and ‘j’ are indexes of particles and N is particle number. For Eq. (A4), the momentum density conservation law

$$\begin{aligned} \frac{\partial J(\vec {\mathbf{r}},t)}{\partial t } = -\nabla _\mathbf{r} \cdot \vec {P}, \end{aligned}$$

(A5)

is needed. Moreover it needs

$$\begin{aligned}&\frac{\partial }{\partial \vec {\mathbf{r}}_{ {i}} } \delta (\vec {\mathbf{r}}-\vec {\mathbf{r}}_{i})= - \frac{\partial }{\partial \vec {\mathbf{r}} } \delta (\vec {\mathbf{r}}-\vec {\mathbf{r}}_{i}), \end{aligned}$$

(A6)

$$\begin{aligned}&\delta (\vec {\mathbf{r}}-\vec {\mathbf{r}}_{i}) -\delta (\vec {\mathbf{r}}-\vec {\mathbf{r}}_{j})\approx \vec {R}_{ij}\frac{\partial }{\partial \vec {\mathbf{r}} } \delta (\vec {\mathbf{r}}-\vec {\mathbf{r}}_{j}). \end{aligned}$$

(A7)

One can define the Fourier transform and inverse in three-dimensions by

$$\begin{aligned}&f(\vec {\mathbf{k}}) = \int d^{3}{} \mathbf{r} f(\vec {\mathbf{r}}) \mathrm{exp} [ \mathrm{i} \vec {\mathbf{k}}\cdot \vec {\mathbf{r}}], \end{aligned}$$

(A8)

$$\begin{aligned}&f(\vec {\mathbf{r}}) = \frac{1}{(2\pi )^{3}}\int d^{3}{} \mathbf{k} f(\vec {\mathbf{k}}) \mathrm{exp} [- \mathrm{i}\vec {\mathbf{k}}\cdot \vec {\mathbf{r}} ], \end{aligned}$$

(A9)

where ‘$\mathrm{i}$’ is the unit of imaginary. Then the mass density, momentum density and stress tensor in $\vec {\mathbf{k}}$-space are given

$$\begin{aligned}&\rho (\vec {\mathbf{k}},t)={\sum _{i}^{N}} {m_{i}} \mathrm{exp} [ \mathrm{i} \vec {\mathbf{k}}\cdot \vec {\mathbf{r}}_{i} ], \end{aligned}$$

(A10)

$$\begin{aligned}&J(\vec {\mathbf{k}},t)=\sum _{i}^{N} m_{i} {\dot{\vec {\mathbf{r}}}_{ {i}}} \mathrm{exp} [ \mathrm{i} \vec {\mathbf{k}}\cdot \vec {\mathbf{r}}_{i} ], \end{aligned}$$

(A11)

$$\begin{aligned} P_{\alpha \beta }(\vec {\mathbf{k}},t)&=\sum _{i}^{N} \frac{p_{{i}\alpha }p_{{i}\beta }}{m_{i}} \mathrm{exp} [ \mathrm{i} \vec {\mathbf{k}}\cdot \vec {\mathbf{r}}_{i} ] \\ \nonumber&+\frac{1}{2} \sum _{i}^{N} \sum _{i\ne j}^{N} F_{ij\alpha }R_{ij\beta } \mathrm{exp} [ \mathrm{i} \vec {\mathbf{k}}\cdot \vec {\mathbf{r}}_{j} ]+\ldots \; . \end{aligned}$$

(A12)

A.2 Shear viscosity and strain rate

The stress tensor corresponds to the strain rate (for simplicity, only the $x-y$ component is considered) reads

$$\begin{aligned} {\begin{matrix} P_{xy}=-\eta \gamma (t), \end{matrix}} \end{aligned}$$

(A13)

where $\eta$ is static shear viscosity. Also one can see it in Eq. (9) but the strain rate here is time dependent. The most general linear relation between the strain rate and the shear stress can be written in the time domain as

$$\begin{aligned} {\begin{matrix} P_{xy}(t)=-\int _{0}^{t} ds \; \eta _{M}(t-s)\gamma (s), \end{matrix}} \end{aligned}$$

(A14)

where $\eta _{M}(t)$ is called the Maxwell memory function. The memory function explains that the shear stress at time t is not simply linearly proportional to the strain rate at the current time t, but to the entire strain rate process, over times $0 \leqslant s \leqslant t$. For the frequency dependent Maxwell viscosity is

$$\begin{aligned} \tilde{\eta }_{M}(\omega )= \frac{{\eta }}{1+{\mathrm{i}} \omega \tau _{M}}, \end{aligned}$$

(A15)

where $\tau _{M}$ is the Maxwell relaxation time which controls the transition frequency between low frequency viscous behaviour and high frequency elastic behavior. In Eq. (A15), $\tilde{\eta }_{M}$ is the Fourier–Laplace transform which is read as

$$\begin{aligned} \tilde{\eta }(\omega )= {\int _{0}^{\infty }} dt\, {\hbox {exp}}{[}-{\mathrm{i}}\omega t{]} \eta (t). \end{aligned}$$

(A16)

A.3 Shear viscosity of the Green–Kubo formula

We separate vector-dependent momentum density into longitudinal ($\mathbf{J}^{||}$) and transverse ($\mathbf{J}^{\bot }$) parts. Considering a transverse momentum density $\mathbf{J}^{\bot }$($\vec {\mathbf{k}}$,t), for simplicity, we define the coordinate system in which $\vec {\mathbf{k}}$ is in y-direction and $\mathbf{J}^{\bot }$ is in the x-direction:

$$\begin{aligned} J_{x}(k_{y},t) = \sum _{i}mv_{xi}(t)\mathrm{exp}[\mathrm{i}k_{y}y_{i}(t)]. \end{aligned}$$

(A17)

According to Eq. (A12), one can get

$$\begin{aligned} \dot{J}_{x}(k_{y},t) = \mathrm{i}k_{y}P_{xy}(k_{y},t). \end{aligned}$$

(A18)

In Ref. [64], by Mori–Zwanzig formalism, one can get the shear viscosity $\eta (t)$ which is time-dependent, i.e.

$$\begin{aligned} \eta (t) = \frac{VNm}{\langle J_{x}(k_{y})J_{x}^{*}(k_{y})\rangle }\langle P_{xy}(t)P_{xy}(0) \rangle . \end{aligned}$$

(A19)

By the Fourier–Laplace transform of $\eta (t)$, one gets

$$\begin{aligned} \tilde{\eta }(\omega ) = \frac{VNm}{\langle J_{x}(k_{y}=0)J_{x}^{*}(k_{y}=0)\rangle } \\ \times \int _{0}^{\infty }\langle P_{xy}(t)P_{xy}(0) \rangle \mathrm{exp}[-\mathrm{i}\omega t]dt.\nonumber \end{aligned}$$

(A20)

As in Eq. (A15), static shear viscosity needs $\omega \rightarrow$0. Then one gets

$$\begin{aligned} \eta = \frac{VNm}{\langle J_{x}(k_{y} = 0) J_{x}^{*}(k_{y} = 0)\rangle } \int _{0}^{\infty }\langle P_{xy}(t)P_{xy}(0) \rangle dt. \end{aligned}$$

(A21)

Here it should be noticed

$$\begin{aligned} P_{xy}(t)&= \lim _{k_{y}\rightarrow 0}\frac{P_{xy}(k_{y},t) }{V}, \end{aligned}$$

(A22)

$$\begin{aligned} P_{xy}(0)&= \lim _{k_{y}\rightarrow 0}\frac{P_{xy}(k_{y},0) }{V}, \end{aligned}$$

(A23)

where V is system volume. For the norm of the transverse current, one can get

$$\begin{aligned}&\langle J_{x}(k_{y}=0) J_{x}^{*}(k_{y}=0)\rangle \\&=\left\langle \sum _{i}^{N}p_{xi} \sum _{j}^{N}p_{xj}\right\rangle \nonumber \\&=\left\langle \sum _{i}^{N}p_{xi}^{2}\rangle +N(N-1)\langle p_{1x}p_{2x}\right\rangle . \nonumber \end{aligned}$$

(A24)

At equilibrium, $p_{1x}$ is independent of $p_{2x}$, so the second term of right-hand side of Eq. (A24) is zero. Then we can obtain a new form of the Green–Kubo formula for shear viscosity

$$\begin{aligned} \eta _{new} = \frac{VNm}{\left\langle \sum _{i}^{N}p_{xi}^{2}\right\rangle } \int _{0}^{\infty }\langle P_{xy}(t)P_{xy}(0) \rangle dt, \end{aligned}$$

(A25)

where $\langle \ldots \rangle$ denotes ensemble average. For an equilibrium system which obeys the Boltzmann distribution, one can get

$$\begin{aligned} \left\langle \sum _{i}^{N}p_{xi}^{2} \right\rangle = \left\langle \frac{1}{3}\sum _{i}^{N}p_{i}^{2} \right\rangle = Nmk_{B}T, \end{aligned}$$

(A26)

where T is temperature and $k_{B}$ ($k_{B}$ = 1) is the Boltzmann constant. Then the normal Green–Kubo formula for shear viscosity can be given

$$\begin{aligned} \eta _{nor} = \frac{V}{T} \int _{0}^{\infty }\langle P_{xy}(t)P_{xy}(0) \rangle dt. \end{aligned}$$

(A27)